3.262 \(\int \cosh (\frac {a+b x}{c+d x}) \, dx\)

Optimal. Leaf size=101 \[ \frac {\sinh \left (\frac {b}{d}\right ) (b c-a d) \text {Chi}\left (\frac {b c-a d}{d (c+d x)}\right )}{d^2}-\frac {\cosh \left (\frac {b}{d}\right ) (b c-a d) \text {Shi}\left (\frac {b c-a d}{d (c+d x)}\right )}{d^2}+\frac {(c+d x) \cosh \left (\frac {a+b x}{c+d x}\right )}{d} \]

[Out]

(d*x+c)*cosh((b*x+a)/(d*x+c))/d-(-a*d+b*c)*cosh(b/d)*Shi((-a*d+b*c)/d/(d*x+c))/d^2+(-a*d+b*c)*Chi((-a*d+b*c)/d
/(d*x+c))*sinh(b/d)/d^2

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Rubi [A]  time = 0.18, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5608, 3297, 3303, 3298, 3301} \[ \frac {\sinh \left (\frac {b}{d}\right ) (b c-a d) \text {Chi}\left (\frac {b c-a d}{d (c+d x)}\right )}{d^2}-\frac {\cosh \left (\frac {b}{d}\right ) (b c-a d) \text {Shi}\left (\frac {b c-a d}{d (c+d x)}\right )}{d^2}+\frac {(c+d x) \cosh \left (\frac {a+b x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[(a + b*x)/(c + d*x)],x]

[Out]

((c + d*x)*Cosh[(a + b*x)/(c + d*x)])/d + ((b*c - a*d)*CoshIntegral[(b*c - a*d)/(d*(c + d*x))]*Sinh[b/d])/d^2
- ((b*c - a*d)*Cosh[b/d]*SinhIntegral[(b*c - a*d)/(d*(c + d*x))])/d^2

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5608

Int[Cosh[((e_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_))]^(n_.), x_Symbol] :> -Dist[d^(-1), Subst[Int[Cosh[(
b*e)/d - (e*(b*c - a*d)*x)/d]^n/x^2, x], x, 1/(c + d*x)], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && NeQ[b*
c - a*d, 0]

Rubi steps

\begin {align*} \int \cosh \left (\frac {a+b x}{c+d x}\right ) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\cosh \left (\frac {b}{d}-\frac {(b c-a d) x}{d}\right )}{x^2} \, dx,x,\frac {1}{c+d x}\right )}{d}\\ &=\frac {(c+d x) \cosh \left (\frac {a+b x}{c+d x}\right )}{d}+\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {b}{d}-\frac {(b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{d^2}\\ &=\frac {(c+d x) \cosh \left (\frac {a+b x}{c+d x}\right )}{d}-\frac {\left ((b c-a d) \cosh \left (\frac {b}{d}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {(b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{d^2}+\frac {\left ((b c-a d) \sinh \left (\frac {b}{d}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {(b c-a d) x}{d}\right )}{x} \, dx,x,\frac {1}{c+d x}\right )}{d^2}\\ &=\frac {(c+d x) \cosh \left (\frac {a+b x}{c+d x}\right )}{d}+\frac {(b c-a d) \text {Chi}\left (\frac {b c-a d}{d (c+d x)}\right ) \sinh \left (\frac {b}{d}\right )}{d^2}-\frac {(b c-a d) \cosh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {b c-a d}{d (c+d x)}\right )}{d^2}\\ \end {align*}

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Mathematica [B]  time = 0.37, size = 373, normalized size = 3.69 \[ \frac {(b c-a d) \left (\sinh \left (\frac {b}{d}\right )-\cosh \left (\frac {b}{d}\right )\right ) \text {Chi}\left (\frac {b c-a d}{x d^2+c d}\right )+(b c-a d) \left (\sinh \left (\frac {b}{d}\right )+\cosh \left (\frac {b}{d}\right )\right ) \text {Chi}\left (\frac {a d-b c}{d (c+d x)}\right )-a d \sinh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {b c-a d}{x d^2+c d}\right )+b c \sinh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {b c-a d}{x d^2+c d}\right )+a d \cosh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {b c-a d}{x d^2+c d}\right )-b c \cosh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {b c-a d}{x d^2+c d}\right )+2 d^2 x \cosh \left (\frac {a+b x}{c+d x}\right )-a d \sinh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {a d-b c}{d (c+d x)}\right )+b c \sinh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {a d-b c}{d (c+d x)}\right )-a d \cosh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {a d-b c}{d (c+d x)}\right )+b c \cosh \left (\frac {b}{d}\right ) \text {Shi}\left (\frac {a d-b c}{d (c+d x)}\right )+2 c d \cosh \left (\frac {a+b x}{c+d x}\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[(a + b*x)/(c + d*x)],x]

[Out]

(2*c*d*Cosh[(a + b*x)/(c + d*x)] + 2*d^2*x*Cosh[(a + b*x)/(c + d*x)] + (b*c - a*d)*CoshIntegral[(b*c - a*d)/(c
*d + d^2*x)]*(-Cosh[b/d] + Sinh[b/d]) + (b*c - a*d)*CoshIntegral[(-(b*c) + a*d)/(d*(c + d*x))]*(Cosh[b/d] + Si
nh[b/d]) + b*c*Cosh[b/d]*SinhIntegral[(-(b*c) + a*d)/(d*(c + d*x))] - a*d*Cosh[b/d]*SinhIntegral[(-(b*c) + a*d
)/(d*(c + d*x))] + b*c*Sinh[b/d]*SinhIntegral[(-(b*c) + a*d)/(d*(c + d*x))] - a*d*Sinh[b/d]*SinhIntegral[(-(b*
c) + a*d)/(d*(c + d*x))] - b*c*Cosh[b/d]*SinhIntegral[(b*c - a*d)/(c*d + d^2*x)] + a*d*Cosh[b/d]*SinhIntegral[
(b*c - a*d)/(c*d + d^2*x)] + b*c*Sinh[b/d]*SinhIntegral[(b*c - a*d)/(c*d + d^2*x)] - a*d*Sinh[b/d]*SinhIntegra
l[(b*c - a*d)/(c*d + d^2*x)])/(2*d^2)

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fricas [A]  time = 0.52, size = 171, normalized size = 1.69 \[ \frac {2 \, {\left (d^{2} x + c d\right )} \cosh \left (\frac {b x + a}{d x + c}\right ) - {\left ({\left (b c - a d\right )} {\rm Ei}\left (\frac {b c - a d}{d^{2} x + c d}\right ) - {\left (b c - a d\right )} {\rm Ei}\left (-\frac {b c - a d}{d^{2} x + c d}\right )\right )} \cosh \left (\frac {b}{d}\right ) + {\left ({\left (b c - a d\right )} {\rm Ei}\left (\frac {b c - a d}{d^{2} x + c d}\right ) + {\left (b c - a d\right )} {\rm Ei}\left (-\frac {b c - a d}{d^{2} x + c d}\right )\right )} \sinh \left (\frac {b}{d}\right )}{2 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((b*x+a)/(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*(d^2*x + c*d)*cosh((b*x + a)/(d*x + c)) - ((b*c - a*d)*Ei((b*c - a*d)/(d^2*x + c*d)) - (b*c - a*d)*Ei(-
(b*c - a*d)/(d^2*x + c*d)))*cosh(b/d) + ((b*c - a*d)*Ei((b*c - a*d)/(d^2*x + c*d)) + (b*c - a*d)*Ei(-(b*c - a*
d)/(d^2*x + c*d)))*sinh(b/d))/d^2

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giac [B]  time = 5.29, size = 764, normalized size = 7.56 \[ \frac {{\left (b^{3} c^{2} {\rm Ei}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\frac {b}{d}} - 2 \, a b^{2} c d {\rm Ei}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\frac {b}{d}} - \frac {{\left (b x + a\right )} b^{2} c^{2} d {\rm Ei}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\frac {b}{d}}}{d x + c} + a^{2} b d^{2} {\rm Ei}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\frac {b}{d}} + \frac {2 \, {\left (b x + a\right )} a b c d^{2} {\rm Ei}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\frac {b}{d}}}{d x + c} - \frac {{\left (b x + a\right )} a^{2} d^{3} {\rm Ei}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\frac {b}{d}}}{d x + c} + b^{2} c^{2} d e^{\left (\frac {b x + a}{d x + c}\right )} - 2 \, a b c d^{2} e^{\left (\frac {b x + a}{d x + c}\right )} + a^{2} d^{3} e^{\left (\frac {b x + a}{d x + c}\right )}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )}}{2 \, {\left (b d^{2} - \frac {{\left (b x + a\right )} d^{3}}{d x + c}\right )}} - \frac {{\left (b^{3} c^{2} {\rm Ei}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\left (-\frac {b}{d}\right )} - 2 \, a b^{2} c d {\rm Ei}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\left (-\frac {b}{d}\right )} - \frac {{\left (b x + a\right )} b^{2} c^{2} d {\rm Ei}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\left (-\frac {b}{d}\right )}}{d x + c} + a^{2} b d^{2} {\rm Ei}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\left (-\frac {b}{d}\right )} + \frac {2 \, {\left (b x + a\right )} a b c d^{2} {\rm Ei}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\left (-\frac {b}{d}\right )}}{d x + c} - \frac {{\left (b x + a\right )} a^{2} d^{3} {\rm Ei}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) e^{\left (-\frac {b}{d}\right )}}{d x + c} - b^{2} c^{2} d e^{\left (-\frac {b x + a}{d x + c}\right )} + 2 \, a b c d^{2} e^{\left (-\frac {b x + a}{d x + c}\right )} - a^{2} d^{3} e^{\left (-\frac {b x + a}{d x + c}\right )}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )}}{2 \, {\left (b d^{2} - \frac {{\left (b x + a\right )} d^{3}}{d x + c}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((b*x+a)/(d*x+c)),x, algorithm="giac")

[Out]

1/2*(b^3*c^2*Ei(-(b - (b*x + a)*d/(d*x + c))/d)*e^(b/d) - 2*a*b^2*c*d*Ei(-(b - (b*x + a)*d/(d*x + c))/d)*e^(b/
d) - (b*x + a)*b^2*c^2*d*Ei(-(b - (b*x + a)*d/(d*x + c))/d)*e^(b/d)/(d*x + c) + a^2*b*d^2*Ei(-(b - (b*x + a)*d
/(d*x + c))/d)*e^(b/d) + 2*(b*x + a)*a*b*c*d^2*Ei(-(b - (b*x + a)*d/(d*x + c))/d)*e^(b/d)/(d*x + c) - (b*x + a
)*a^2*d^3*Ei(-(b - (b*x + a)*d/(d*x + c))/d)*e^(b/d)/(d*x + c) + b^2*c^2*d*e^((b*x + a)/(d*x + c)) - 2*a*b*c*d
^2*e^((b*x + a)/(d*x + c)) + a^2*d^3*e^((b*x + a)/(d*x + c)))*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)/(b*d^2 -
(b*x + a)*d^3/(d*x + c)) - 1/2*(b^3*c^2*Ei((b - (b*x + a)*d/(d*x + c))/d)*e^(-b/d) - 2*a*b^2*c*d*Ei((b - (b*x
+ a)*d/(d*x + c))/d)*e^(-b/d) - (b*x + a)*b^2*c^2*d*Ei((b - (b*x + a)*d/(d*x + c))/d)*e^(-b/d)/(d*x + c) + a^
2*b*d^2*Ei((b - (b*x + a)*d/(d*x + c))/d)*e^(-b/d) + 2*(b*x + a)*a*b*c*d^2*Ei((b - (b*x + a)*d/(d*x + c))/d)*e
^(-b/d)/(d*x + c) - (b*x + a)*a^2*d^3*Ei((b - (b*x + a)*d/(d*x + c))/d)*e^(-b/d)/(d*x + c) - b^2*c^2*d*e^(-(b*
x + a)/(d*x + c)) + 2*a*b*c*d^2*e^(-(b*x + a)/(d*x + c)) - a^2*d^3*e^(-(b*x + a)/(d*x + c)))*(b*c/(b*c - a*d)^
2 - a*d/(b*c - a*d)^2)/(b*d^2 - (b*x + a)*d^3/(d*x + c))

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maple [B]  time = 0.12, size = 347, normalized size = 3.44 \[ \frac {{\mathrm e}^{-\frac {b x +a}{d x +c}} a}{\frac {2 d a}{d x +c}-\frac {2 b c}{d x +c}}-\frac {{\mathrm e}^{-\frac {b x +a}{d x +c}} c b}{2 d \left (\frac {d a}{d x +c}-\frac {b c}{d x +c}\right )}-\frac {{\mathrm e}^{-\frac {b}{d}} \Ei \left (1, \frac {d a -c b}{d \left (d x +c \right )}\right ) a}{2 d}+\frac {{\mathrm e}^{-\frac {b}{d}} \Ei \left (1, \frac {d a -c b}{d \left (d x +c \right )}\right ) c b}{2 d^{2}}+\frac {d \,{\mathrm e}^{\frac {b x +a}{d x +c}} x a}{2 d a -2 c b}-\frac {{\mathrm e}^{\frac {b x +a}{d x +c}} x c b}{2 \left (d a -c b \right )}+\frac {{\mathrm e}^{\frac {b x +a}{d x +c}} c a}{2 d a -2 c b}-\frac {{\mathrm e}^{\frac {b x +a}{d x +c}} c^{2} b}{2 d \left (d a -c b \right )}+\frac {{\mathrm e}^{\frac {b}{d}} \Ei \left (1, -\frac {d a -c b}{d \left (d x +c \right )}\right ) a}{2 d}-\frac {{\mathrm e}^{\frac {b}{d}} \Ei \left (1, -\frac {d a -c b}{d \left (d x +c \right )}\right ) c b}{2 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh((b*x+a)/(d*x+c)),x)

[Out]

1/2*exp(-(b*x+a)/(d*x+c))/(d*a/(d*x+c)-b*c/(d*x+c))*a-1/2/d*exp(-(b*x+a)/(d*x+c))/(d*a/(d*x+c)-b*c/(d*x+c))*c*
b-1/2/d*exp(-b/d)*Ei(1,(a*d-b*c)/d/(d*x+c))*a+1/2/d^2*exp(-b/d)*Ei(1,(a*d-b*c)/d/(d*x+c))*c*b+1/2*d*exp((b*x+a
)/(d*x+c))/(a*d-b*c)*x*a-1/2*exp((b*x+a)/(d*x+c))/(a*d-b*c)*x*c*b+1/2*exp((b*x+a)/(d*x+c))/(a*d-b*c)*c*a-1/2/d
*exp((b*x+a)/(d*x+c))/(a*d-b*c)*c^2*b+1/2/d*exp(b/d)*Ei(1,-(a*d-b*c)/d/(d*x+c))*a-1/2/d^2*exp(b/d)*Ei(1,-(a*d-
b*c)/d/(d*x+c))*c*b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cosh \left (\frac {b x + a}{d x + c}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((b*x+a)/(d*x+c)),x, algorithm="maxima")

[Out]

integrate(cosh((b*x + a)/(d*x + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {cosh}\left (\frac {a+b\,x}{c+d\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh((a + b*x)/(c + d*x)),x)

[Out]

int(cosh((a + b*x)/(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cosh {\left (\frac {a + b x}{c + d x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh((b*x+a)/(d*x+c)),x)

[Out]

Integral(cosh((a + b*x)/(c + d*x)), x)

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